Semi-direct decompositions of Banach algebras Let $A$ be a Banach algebra and I an ideal of it. Is there always a subalgebra $B$ of $A$ such that A can be written as $A=B\oplus I$ where $\oplus$ denotes direct sum?
If not, in what conditions we have such decomposition? (conditions on A and I).
P.S. I am new to this subject. Please give some references (papers or books) that I can find something related to such decompositions. Thanks.
 A: For example, consider $A = C[0,1]$, the continuous real-valued (or complex- if you prefer) functions on $[0,1]$, and $I = \{f \in A: f(0) = f(1) = 0\}$.  If $g \in B$ we have
$h = g^3 - (g(0) + g(1)) g^2 + g(0) g(1) g \in B \cap I$, and $h \ne 0$ unless $g$ is constant.
The counterexample generalizes somewhat: suppose $A$ is a semisimple commutative Banach algebra over $\mathbb C$ with a connected maximal ideal space and $I = I_1 \cap I_2$ where $I_1$ and $I_2$ are two distinct maximal ideals of $A$, corresponding to characters $\chi_1$ and $\chi_2$.  For any $g \in B$ we have
$h = g^3 - (\chi_1(g) + \chi_2(g)) g^2 + \chi_1(g) \chi_2(g) g \in B \cap I$
so $h = 0$.  But then for any character $\chi$, $0 = \chi(h) = (\chi(g)-\chi(g_1)) (\chi(g) - \chi(g_2)) \chi(g)$ so either $\chi(g) = \chi(g_1)$ or $\chi(g) = \chi(g_2)$ or $\chi(g) = 0$.  Since the maximal ideal space is connected, the Gelfand representation $\widehat{g}$ is constant.  But this can't be the case since $B$ must be two-dimensional and the Gelfand representation is one-to-one.
A: Take $A=C([0,1])$, and let $I=\{g\in A\colon g(0)=g(1)=0\}$. If $f\in B$ and $g\in I$ with $f(x)+g(x)=x$ for $x\in[0,1]$ then $f(0)=0$ and $f(1)=1$, and $f-f^2\in B\cap I$. But yet $f-f^2\ne0$.
A: There are already two very good answers but let me add one more. 
One in general cannot expect such a result because $A$, regarded only as a Banach space, can have very few complemented subspaces. Indeed, Koszmider constructed compact, Hausdroff spaces such that the only complemented subspaces of $C(K)$ are the finite- and cofinite-dimensional subspaces (namely the trivial ones). See

Piotr Koszmider, A survey on Banach spaces $C(K)$ with few operators, RACSAM 104 (2), 2010, 309-326.

In particular, if $A=C(K)$ as above, then the only ideals which admit such a decomposition are $\{0\}$ and $C(K)$ itself.
